How to figure out the bounds of double integration? I have a double integration problem that I do not understand how the bounds are what they are. The question is:

Find the volume of the given solid under the surface $z = 5xy$ and
  above the triangle with vertices  $(1, 1), (4, 1),$ and $(1, 2)$.

Why are the bounds for this problem $(1 \leq y \leq 2$ and $1 \leq x \leq 7-3y)$ instead of $(7-3y \leq x \leq y-1$ and $1 \leq y \leq 4)$?
 A: One way to find the bounds is to use the vertices of the triangle to help you. The hypotenuse of the triangle is defined by the segment connecting $(4,1)$ and $(1,2)$ if you find the equation of the line that correspond to this segment you can get a better idea of what the bounds are. Since you have a simple shape (a right triangle), you can set either the x or y variables to be the independent variable. Let's take the y variable as the independent variable.
Start by finding the slope of the segment
$$\left(\frac{1-2}{4-1}\right) = \frac{-1}{3}$$
Then, you can use the point $(1,1)$ and the slope intercept form formula to find the equation of the line, which results in this:
$$ y = \left(\frac{-1}{3}\right)x + \left(\frac{2}{3}\right) $$
Solving for x gives:
$$ x = 2-3y $$
This equation is dependent on y and determines how far you can go on the x direction. 
The upper bound is the equation and the lower bound is the smallest x value in your set of points. Thus,
$$(1 \leq x \leq 2-3y)$$
Since y was made to be the independent variable, all you need to do is find the smallest and largest y values in your set of points. So, using $(1,1)$ and $(1,2)$ we can tell that the y bounds are $$1 \leq y \leq 2$$
